Homework III CJ

Physics III

Homework III CJ

Chapter 27; 76

Chapter 28; 82

Chapter 29; 7, 14, 20, 30, 50, 56, 61

27.76.Identify: The torque exerted by the magnetic field is The torque required to hold the loop in place is

Set Up: is normal to the plane of the loop, with a direction given by the right-hand rule that is illustrated in Figure 27.32 in the textbook. where is the angle between the normal to the loop and the direction of

Execute:
(a) , in the direction. To keep the loop in place, you must provide a torque in the direction.

(b) in the direction. You must provide a torque in the direction to keep the loop in place.

Evaluate:
(c) If the loop was pivoted through its center, then there would be a torque on both sides of the loop parallel to the rotation axis. However, the lever arm is only half as large, so the total torque in each case is identical to the values found in parts (a)and(b).

28.82.Identify: Find the vector sum of the fields due to each sheet.

Set Up: Problem 28.81 shows that for an infinite sheet . If I is out of the page, is to the left above the sheet and to the right below the sheet. If I is into the page, is to the right above the sheet and to the left below the sheet. B is independent of the distance from the sheet. The directions of the two fields at points P, R and S are shown in Figure 28.82.

Execute:
(a) Above the two sheets, the fields cancel (since there is no dependence upon the distance from the sheets).

(b) In between the sheets the two fields add up to yield to the right.

Evaluate: The two sheets with currents in opposite directions produce a uniform field between the sheets and zero field outside the two sheets. This is analogous to the electric field produced by large parallel sheets of charge of opposite sign.

Figure 28.82

29.7.Identify: Calculate the flux through the loop and apply Faraday’s law.

Set Up: To find the total flux integrate over the width of the loop. The magnetic field of a long straight wire, at distance r from the wire, is . The direction of is given by the right-hand rule.

Execute:
(a) When , into the page.

(b)

(c)

(d)

(e)

Evaluate: The induced emf is proportional to the rate at which the current in the long straight wire is changing

29.14.Identify: A change in magnetic flux through a coil induces an emf in the coil.

Set Up: The flux through a coil is  = NBAcos and the induced emf is

Execute: The flux is constant in each case, so the induced emf is zero in all cases.

Evaluate: Even though the coil is moving within the magnetic field and has flux through it, this flux is not changing, so no emf is induced in the coil.

29.20.Identify: Use the results of Example 29.6. Use the three approaches specified in the problem for determining the direction of the induced current. .

Set Up: Let be directed into the figure, so a clockwise emf is positive.

Execute:
(a)

(b)
(i) Let q be a positive charge in the moving bar, as shown in Figure 29.20a. The magnetic force on this charge is , which points upward. This force pushes the current in a counterclockwise direction through the circuit.

(ii) is positive and is increasing in magnitude, so . Then by Faraday’s law and the emf and induced current are counterclockwise.

(iii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase. Hence this current must flow in a counterclockwise sense, as shown in Figure 29.20b.

(c)

Evaluate: All three methods agree on the direction of the induced current.

Figure 29.20

29.30.Identify: Use Eq.(29.10) to calculate the induced electric field E and use this E in Eq.(29.9) to calculate between two points.

(a) Set Up: Becauseoftheaxialsymmetryandtheabsenceofanyelectriccharge,thefieldlinesareconcentriccircles.
(b) See Figure 29.30.

/ is tangent to the ring. The direction of (clockwise or counterclockwise) is the direction in which current will be induced in the ring.
Figure 29.30

Execute: Use the sign convention for Faraday’s law to deduce this direction. Let be into the paper. Then is positive. B decreasing then means is negative, so by is positive and therefore clockwise. Thus is clockwise around the ring. To calculate E apply to a circular path that coincides with the ring.

(d) Points a and b are separated by a distance around the ring of so

(e) The ends are separated by a distance around the ring of so as calculated in part (c).

Evaluate: The induced emf, calculated from Faraday’s law and used to calculate the induced current, is associated with the induced electric field integrated around the total circumference of the ring.

29.50.Identify: Apply Faraday’s law.

Set Up: For rotation about the y-axis the situation is the same as in Examples 29.4 and 29.5 and we can apply the results from those examples.

Execute:
(a) Rotating about the y-axis: the flux is given by and

(b) Rotating about the x-axis: and

Evaluate: The maximum emf is the same if the loop is rotated about an edge parallel to the z-axis as it is when it is rotated about the z-axis.

29.56.Identify: Apply Newton’s 2nd law to the bar. The bar will experience a magnetic force due to the induced current in the loop. Use to solve for v. At the terminal speed, .

Set Up: The induced emf in the loop has a magnitude. The induced emf is counterclockwise, so it opposes the voltage of the battery,

Execute:
(a) The net current in the loop is The acceleration of the bar is To find , set and solve for using the method of separation of variables:

The graph of v versus t is sketched in Figure 29.56. Note that the graph of this function is similar in appearance to that of a charging capacitor.

(b) Just after the switch is closed, and and

(d) Note that as the speed increases, the acceleration decreases. The speed will asymptotically approach the terminal speed which makes the acceleration zero.
Evaluate: The current in the circuit is counterclockwise and the magnetic force on the bar is to the right. The energy that appears as kinetic energy of the moving bar is supplied by the battery.

Figure 29.56

29.61.(a) and (b) Identifyand Set Up:

/ The magnetic field of the wire is given by and varies along the length of the bar. At every point along the bar has direction into the page. Divide the bar up into thin slices, as shown in Figure 29.61a.
Figure 29.61a

Execute: The emf induced in each slice is given by is directed toward the wire, so The total emf induced in the bar is

Evaluate: The minus sign means that is negative, point a is at higher potential than point b. (The force on positive charge carriers in the bar is towards a, so a is at higher potential.) The potential difference increases when I or v increase, or d decreases.

Set Up: The wire and loop are sketched in Figure 29.61b.

/ Execute: As the loop moves to the right the magnetic flux through it doesn’t change. Thus and I = 0.
Figure 29.61b

Evaluate: This result can also be understood as follows. The induced emf in section ab puts point a at higher potential; the induced emf in section dc puts point d at higher potential. If you travel around the loop then these two induced emf’s sum to zero. There is no emf in the loop and hence no current.